# Prove a metric space is totally bounded iff it is bounded in every equivalent metric.

Prove a metric space is totally bounded iff it is bounded in every equivalent metric.

I was able to see the solution to this problem if we change the wording to "iff it is totally bounded in every equivalent metric." I was wondering whether there is a typo in the wording because bounded does not imply totally bounded, and just because the metric space is bounded in equivalent metric does not give much information as to whether space is totally bounded.

• What is your definition of an equivalent metric? There are two commonly used (and inequivalent) definitions, see here. Apr 9, 2021 at 8:24
• the definition im using is that open ball in one metric can be contained in open ball in the other metric and vice versa. if that is the case, then the metrics are equivalent
– Bill
Apr 9, 2021 at 8:30
• OK, in other words, for you, metrics are equivalent iff they define the same topology. Then indeed, you are asked to prove a wrong claim. Most likely, though, whoever wrote the problem had in mind a different definition of equivalence. Apr 9, 2021 at 8:40
• @MoisheKohan The statement is even more blatantly false with the other definition. Becasue it would say that a metric space is totalyl bounded iff it is bounded! Apr 9, 2021 at 8:46
• @MoisheKohan yes its definitely a typo
– Bill
Apr 9, 2021 at 9:58

This is false. $$(0,1)$$ is totally bounded w.r.t. the usual metric. An equivalent metric is $$|\frac 1 x-\frac 1 y|$$ and $$(0,1)$$ is not bounded in this metric.

[Definition of equivalent metrics I am using: two metrics are equivalent if they have the same convergent sequeneces with the same limits; equivalently, they have the same open sets].

1. There was a typo and the problem should read

Prove a metric space is totally bounded iff it is totally bounded in every equivalent metric.

and

1. The definition of "equivalent" in the problem was not completely standard, the author of the problem was assuming strong equivalence of metrics. (Or, more generally, that the identity map between the two metric spaces $$(X,d_1)\to (X,d_2)$$ is a uniform homeomorphism.)

With these two corrections, the problem becomes a pleasant exercise.

Addendum. Another possibility is that the question was meant to be:

Prove that a metric space is totally bounded if and only if it is bounded in every uniformly equivalent metric.

Here two metrics $$d_1, d_2$$ on a set $$X$$ are called uniformly equivalent if both identity maps $$id: (X, d_1)\to (X,d_2), id: (X, d_2)\to (X,d_1)$$ are uniformly continuous.

This question again has a positive answer but a proof requires much more work. Since the question is likely used as a homework or an exam problem, I will refrain from writing a proof.

• @Gae.S.: Of course, as I said, I am assuming that there was a typo and strong (or uniform) equivalence was assumed. Then the claim holds and a proof is not that hard. I am reluctant to give more than hints since this might be a homework problem. Apr 9, 2021 at 11:43
• Ah, ok, I see your point now (I was misreading). I had been thinking for a bit about the same issue, but with (1) and (2) being mutually exclusive, so I thought you were making a mistake I had already made. I wasn't suggesting that you should add details.
– user239203
Apr 9, 2021 at 11:50